The convolution theorem states that the Fourier transform of a convolution is the pointwise product of Fourier transforms. For a convolution given by

\begin{align} y (x',y',t') &= \displaystyle \int_{t}\int_{y}\int_{x} f(x,y,t) g(x'- x, y'-y, t'-t) \mathrm{d}x\;\mathrm{d}y\;\mathrm{d}t \\ &= f \otimes g \end{align}

we can write

\begin{align} \tilde{y} (k_x,k_y,\omega) &= \tilde{f} (k_x,k_y,\omega) \tilde{g} (k_x,k_y,\omega) \end{align}

where \(\tilde{y}, \tilde{f}\) and \(\tilde{g}\) are the Fourier transform of \(y,f\) and \(g\), respectivly.

Note that the convolution is commutative, meaning that

\begin{align} f \otimes g = g \otimes f \end{align}

Because of that the convolution theorem is independent of which function is reversed and shifted in the convolution integral.